Q. 164.1( 16 Votes )

The angle of elevation of the top of a tower from a point A on the ground is 30°. On moving a distance of 20 metres towards the foot of the tower to a point B the angle of elevation increases to 60°. Find the height of the tower and the distance of the tower from the point A.

Answer :

Let the height of the tower is = h (m)

Distance of point B from foot of the tower is = (m)

In ∆ADC,

tan 30° =


√3 h = 20+ -----------(1)

In ∆DCB,

tan 60° =

√3 =

h = √3 ----------(2)

On substituting value of h from eqn. (2) in eqn. (1)

√3× √3 = 20 +


3-= 20

= 10

Therefore distance of point A from tower is


AC = 20+10 30

Ac = 30 m.

Now substituting value of in eqn. (1)

h = 20+10 30

17.32 m.

Therefore height of tower is 17.32 m.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Trick to learn all Trigonometric Formulae28 mins
Champ Quiz | Trigger on Trigonometry47 mins
Fundas of Trigonometry50 mins
Champ Quiz | Trigonometry Important Questions33 mins
NCERT | Trigonometric Identities52 mins
Champ Quiz | NTSE Trigonometry50 mins
Testing the T- Ratios of Specified Angles57 mins
Foundation | Cracking Previous Year IMO QuestionsFREE Class
NCERT | Imp. Qs. on Trigonometry42 mins
Quiz | Trail of Mixed Questions on Trigonometry59 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses